Math 455                      Homework                      Saff&Snider

Sec 4.3, p128
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#1e. (sin z)^2 cos z has an antiderivative (1/3) (sin z)^3.

#1f. e^z cos z should be integrated by parts.

#1g. z^(1/2) has an antiderivative (2/3) z^(3/2).

#1h. Note that integrating by parts provides us with an antiderivative 
     for (Log z)^2.  Namely,

     F(z) = z (Log z)^2 - 2z (Log z - 1) is an antiderivative for (Log z)^2.

     Hence, using our path independence thm we can simply evaluate this
     antiderivative at the end points of the given path gamma and get
     the value of the integral.  Namely,

     \int_gamma (Log z)^2 d = F(i) - F(1) = (2+pi) + i(2 - pi^2 /4).

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#4.  False.  Note that according to Cauchy's Thm f has to be analytic
     on and inside the closed contour gamma.  Thus, f being analytic
     only on gamma is not sufficient.

     Ex.  The integral of f(z) = 1/z on the closed contour |z|=1.
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#10. We need to show the following (by the def of a lim)

     For any \epsilon > 0 there exists a \delta > 0 s.t.

     |\delta z - 0| < \delta  =>  |int_{0}^{1}f(z+t\delta z) dt - f(z)| <
     \epsilon.

     Now we for an arbitrary \epsilon > 0 we calculate \delta.

     |\int_{0}^{1}f(z+t\delta z) dt - f(z)| = |[\int_0}^{1} f(z + t\delta z)

     - f(z)] dt

     \leq \int_{0}^{1} |f(z+t\delta z) - f(z)| dt, (By Jensen's inequality)

     < \epsilon \int_{0}^{1} dt

     < \epsilon

     since it follows from the continuity of f that 

     for any \epsilon > 0 there is a \delta > 0 s.t.

     |delta z| < \delta  =>  |f(z + t \delta z) - f(z)| < \epsilon.
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#11. Here just follow the hint, i.e., apply Thm 6 to the function d(fg)/dz.
     Consider a contour \gamma. Then, the function d(fg)/dz obviously has
     an antiderivative, fg, in the region of consideration D (f and g must
     have continuous derivative in D which contains \gamma). Hence,

     \int_{\gamma} d(fg)\dz = (fg)(z_2) - (fg)(z_1).

     But, the integrand in the LHS of the above equation can be re-written
     using the product rule for differentiation. That will give the result
     we need.